If \(|\vec{a}| = 3\), \(|\vec{b}| = 5\) and the angle between \(\vec{a}\) and \(\vec{b}\) is \(\displaystyle\frac{2\pi}{3}\), then \(|\vec{a}+\vec{b}|\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\sqrt{19}\)
Since \(|\vec{a}+\vec{b}|^2 = (\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})\),
\(|\vec{a}+\vec{b}|^2 = |\vec{a}|^2+|\vec{b}|^2+2\,\vec{a}\cdot\vec{b}\)
The scalar product is
\( \displaystyle \vec{a}\cdot\vec{b} = 3\times 5\times\cos\frac{2\pi}{3} = 15\times\left(-\frac{1}{2}\right) = -\frac{15}{2} \)
So
\(|\vec{a}+\vec{b}|^2 = 9+25-15 = 19\)
\(|\vec{a}+\vec{b}| = \sqrt{19}\)
Why the other options are incorrect
Option (1) 7 — Incorrect. \(\sqrt{9+25+15} = 7\) uses \(\cos\theta = +\frac{1}{2}\), which is the value for the angle \(\frac{\pi}{3}\).
Option (3) \(\sqrt{34}\) — Incorrect. \(\sqrt{9+25}\) leaves out the \(2\,\vec{a}\cdot\vec{b}\) term; it holds only for perpendicular vectors.
Option (4) 2 — Incorrect. \(5-3 = 2\) would be the value only if the vectors were opposite (angle \(\pi\)).